如何在 mysqli_query 中使用 $row 结果？

Question

如何在 mysqli_query 中使用 $row 结果？

$result1 = mysqli_query($connect,"SELECT subcat_ID FROM subcategories WHERE cat_ID=$cat_ID");
$row=mysqli_fetch_array($result1);
$result2 = mysqli_query($connect,"SELECT subsubcat_name FROM subsubcategories WHERE subcat_ID='".$row['subcat_ID']."'");
while ($row = mysqli_fetch_array($result2)){
        if (isset($row)){
        echo $row['subsubcat_name'];
        echo "<br>";
        }

我没有收到任何错误，但我无法将其打印出来。

知道我做错了什么吗？

感谢您的帮助！

================================================ ====================================

第一次编辑

因为我的问题还没有解决，我再深入一点解释一下：

我的数据库： - 类别：cat_ID、cat_name - 子类别：subcat_ID、cat_ID、extra_cat_ID、subcat_name - 子类别：subsubcat_ID、subcat_ID、subsubcat_name

在某个时刻，我通过 url 获得 cat_ID（类别）的值。 我想要做的是类别的cat_id与子类别中的cat_id或extra_cat_ID的值相同（我也希望能够打印subcat_name），其中子类别的subcat_ID与子类别的subcat_ID相同--->打印subsubcat_name。

Categories
-------------------------
cat_ID | cat_name
------------------
4      | Baby & Kids
5      | Bicycles
6      | Boats
7      | Books & Comics
....
13     | Clothes & Accessories
....
35     | Sport & Fitness
36     | Study
....
38     | Toys & Games
....


Subcategories
-------------------------
subcat_ID | cat_ID | extra_cat_ID | subcat_name
------------------------------------------------
....
15        | 4      | 13           | Baby clothes
16        | 4      | 0            | Baby products
17        | 4      | 13           | Kids clothes
18        | 4      | 38           | Toys
19        | 5      | 0            | Bycicles
20        | 5      | 0            | Bycicle gear & Accessories
21        | 6      | 0            | Boat parts
22        | 6      | 0            | Other Boats
23        | 6      | 0            | Power Boats
24        | 6      | 0            | Sailboats
25        | 6      | 35           | Windsurf & Surfing
26        | 7      | 0            | Antiquarian
27        | 7      | 0            | Books
28        | 7      | 38           | Childrens books
29        | 7      | 0            | Comics
30        | 7      | 0            | Magazines & Newspapers
31        | 7      | 36           | Study & Training



Subsubcategories
-------------------------
subsubcat_ID | subcat_ID | subsubcat_name
-----------------------------------------
...
470          | 15        | Baptism outfits
471          | 15        | Bibs
472          | 15        | Body warmers
473          | 15        | Bodysuits
....
496          | 16        | Baby bath
497          | 16        | Baby books
498          | 16        | Baby inserts
499          | 16        | Baby monitors
....
548          | 17        | Belts
549          | 17        | Blouses & Shirts
550          | 17        | Body warmer
551          | 17        | Boots
....
....
740          | 26        | Music
741          | 26        | Navy
742          | 26        | Novel
743          | 26        | Photography
....
....
867          | 30        | Animals
868          | 30        | Arts and Culture
869          | 30        | Branch
870          | 30        | Cars
870          | 30        | Computers
....
....
etc.

我希望这能澄清一些事情。

Answer 1

您应该使用JOIN，而不是运行两个查询。

例如这个$query：

SELECT subcategories.cat_ID, subsubcategories.subsubcat_name
FROM subcategories
INNER JOIN subsubcategories
ON subcategories.cat_ID=subsubcategories.subcat_ID
WHERE subcategories.cat_ID = ?

在这种情况下，准备查询更合适/更安全：

/* create a prepared statement */
if ($result = mysqli_prepare($connect, $query)) {

    /* bind parameters for markers */
    mysqli_stmt_bind_param($result, "i", $cat_ID);

    /* execute query */
    mysqli_stmt_execute($result);

    /* fetch data */
    while ($row = mysqli_fetch_array($result)){
        if (isset($row)){
            echo $row['subsubcat_name'];
            echo "<br>";
        }
    }
 }

DEMO for SQL JOIN

---

搞定了。这是我使用的代码：

$result=mysqli_query($connect,"SELECT subcategories.cat_ID, subsubcategories.subsubcat_name FROM subcategories INNER JOIN subsubcategories ON subcategories.subcat_ID=subsubcategories.subcat_ID WHERE subcategories.cat_ID = $cat_ID OR subcategories.extra_cat_ID = $cat_ID");
while ($row = mysqli_fetch_array($result)){
        echo $row['subsubcat_name'];
            echo "<br>";
 }

虽然不确定这是安全的方法。随时欢迎提出建议。
感谢 meda 向我展示了另一种方法，但它不是完全相同的代码。

Answer 2

这里是一个如何使用mysqli的例子

面向对象

<?php   
    $connect= new mysqli("localhost","user","passwd","database");
    if ($connect->connect_errno){
      echo "could not connect";
    }

    $select = "SELECT * FROM tablename";

    if($result = $connect->query($select)){
        while($row = $result->fetch_object()){
            echo $row->rowname."<br>";
        }
    }
    else { echo 'no result'; }
    $connect->close();  
?>

程序风格

<?php   
    $connect= mysqli_connect("localhost","user","passwd","database");
    if (mysqli_connect_errno()){
      echo "could not connect";
    }

    $select = "SELECT * FROM tablename";

    if($result = mysqli_query($connect,$select)){
        while($row = mysqli_fetch_object($result)){
            echo $row->rowname."<br>";
        }
    }
    else { echo 'no result'; }
    mysqli_close($connect);  
?>